4 Answers
4

This is a general property of waves. If you have waves reflecting off a clamped point (like waves running on a string that you pinch hard at one point), the waves get phase inverted. The reason is the principle of superposition and the condition that the amplitude at the clamped point is zero. The sum of the reflected and transmitted wave must be the amplitude of oscillation at all points, so that the reflected wave must be phase inverted to cancel the incoming wave.

This property is continuous with the behavior of waves going from a less massive string to a more massive string. The reflection in this case has opposite phase, because the more massive string doesn't respond as quickly to the tension force, and the amplitude of oscillation at the contact point is less than the amplitude of the incoming wave. This means (by superposition) that the reflected wave must cancel part of the incoming wave, and it is phase reflected.

When a wave goes from a more massive string to a less massive string, the less massive string responds with less force, so that the derivative at the oscillating end is flatter than it should be. This means that the reflected wave is reflected in phase with the incoming wave, so that the spatial derivative of the wave is cancelled, not the amplitude reduced.

In optical materials of high density are analogous to strings with a higher density, hence the name. If you go into a material with low speed of light, the time derivative term in the wave-equation is suppressed, so that the field responds more sluggishly, the same way that a massive material responds more sluggishly to tension pulls. Since the eletric field response in these materials is reduced, the reflected wave is phase inverted to make the sum on the surface less, as is appropriate to match with the transmitted wave.

Wave reflections from mismatched impedances have inverted step waves for DC and inverted phases for AC. Just like waves in a pool. :)

added:
Do you equate optically denser to higher relative permitivity to lower relative impedance?
Think of the wavelet as a vector which can only reflect a range of inphase or opposite with null in balance of equal density.

" If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. THis is due to changes in dielectric constant or other physical properties. http://goo.gl/vTwQq

Added:
This illustration should answer your question intuitively with dark bands caused by out of phase or destructive reflection.

Not true. The water wave at the wall of a pool does not get inverted after reflection. It would get inverted if the wall forced a zero of the wave at the boundary, but that's not the case. Waves get inverted in a string with a fixed end, for instance (because there, the wave is forced to be zero).
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TelaclavoMay 9 '12 at 23:36

Transmission Line Theory of light, optical and electromagnetic waves all support what I said about mismatched impedances with the standing wave created by a reflection. I gave the reader's digest description, Perhaps my water example was misleading. Check out Heaviside's transmission line theory based on Maxwell's equations.
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Tony StewartMay 10 '12 at 2:40

I may have explained it poorly.. If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. THis is due to changes in dielectric constant or other physical properties. goo.gl/vTwQq
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Tony StewartMay 10 '12 at 3:01

@Telaclavo close, but not quite. It's not a condition of forced zero, but a condition of forced continuity. The string doesn't have to be forced to be stationary at the discontinuity, but only that the string be rigidly fixed to the second medium (i.e. wall or thicker string). Similarly, the water wave will only reflect inverted if the transverse displacement of the wave was constrained be the same as the transverse displacement of the wall, which is not the case, of course.
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Nathan WiebeMay 11 '12 at 4:13

@NathanWiebe Which sentence that I wrote is not correct? / "Forced continuity" exists almost everywhere (like in all points of a string which are not the ends). Every point of the string is rigidly fixed to the point of string next to it. That does not create any reflection.
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TelaclavoMay 11 '12 at 10:56

Since this has just been asked again, let me attempt an intuitive explanation. The real explanation is of course to match $\vec{E}$ and $\vec{B}$ at the interface and the direction of the reflected wave drops out, but this isn't especially intuitive.

Let's calculate the ratio $E_r/E_i$ as a function of the ratio $n_t/n_i$, and let's start with the refractive indices equal, that is $n_i = n_t$, in which case there is obviously no reflection. As we decrease $n_t/n_i$, either by making $n_t$ smaller or $n_i$ bigger, the reflectivity will increase from zero so we'll get something like (this is the real calculation for the ratio, but the exact form of the graph doesn't matter):

This shows what happens when the refractive index at the incident side is equal to or greater than the refractive index at the transmitted side, but what happens when $n_i < n_t$? Obviously what happens is that we have to continue the line to the left to get something like:

This is the same as the first graph, just continued to values of $n_t/n_i > 1$. The point is that assuming the graph is smooth (which seems physically reasonable) the ratio $E_r/E_i$ must change sign as we pass through $n_t/n_i = 1$. In other words the phase of $E_r$ must differ by $\pi$ on the two sides of the point $n_t/n_i = 1$.

What actually happens is that $\vec{E_i}$ and $\vec{E_r}$ are in phase when $n_t/n_i < 1$ and out of phase by by $\pi$ when $n_t/n_i > 1$, and my argument doesn't prove this. However it hopefully gives you a feel for why the phase of $\vec{E_r}$ must differ (by $\pi$) either side of $n_t/n_i = 1$.